similar to: Substitute NAs by zero

down vote favorite Hello I have a zoo series. It lasts 10 years and its frequency is 15min. I'd like to get a new zoo series (or vector) with the same number of elements, whith each element equal to the first element of the day. That's, The first element everyday is repeated throughout the wole day. This is not same as aggregate(originalseries,as.Date,head,1) because this gives a

Hello I have two series (that can have with different frequencies or with missing values). I merge them and use na.locf, getting a zoo objet with a common index and two core columns. How can I add this columns getting a new zoo series? Any other way of adding two asynchronou series? regards -- View this message in context:

Hello How can I select several not continuous rows ? If I wanted to select rows 1 to 7 I'll write mydata[,1:7] But what if I need to select rows 1 to 5 and 10 to 15? -- View this message in context: http://r.789695.n4.nabble.com/How-to-select-not-continous-rows-tp3003840p3003840.html Sent from the R help mailing list archive at Nabble.com.

Hello Where could I find examples on how to work with the time index in a timeseries or zoo series? Let say I've got this series DATA 1990-01-01 10:00:00 0.900 1990-01-01 10:01:00 0.910 1990-01-01 10:03:00 0.905 1990-01-01 10:04:00 0.905 1990-01-01 10:05:00 0.890 ....................... 2000-12-31 20:00:00 0.992 How do I make simple calculations such as ... ? Calculate the

Hello Imagine I have a vector with ones and zeroes I write it compactly: 1111111100001111111111110000000001111111111100101 I need to get a new vector replacing the "N" ones following the zeroes to new zeroes. For example for N = 3 1111111100001111111111110000000001111111111100101 becomes 1111111100000001111111110000000000001111111100000 I can do it with a for loop but I've read

Hello What's the differente betwen using "plot" and using "print" in order to plot a graph? For example in order to plot the result of a histogram. cheers -- View this message in context: http://r.789695.n4.nabble.com/plot-vs-print-tp3045256p3045256.html Sent from the R help mailing list archive at Nabble.com.

Hello. I don't uderstant when to use textConnection and when not. Some examples do it, some not. I've even seen something like con <- textConnection(rev(rev(ReadLines('data.txt'))[-(1:2])) data <- read.table(con) close(con) -- View this message in context: http://r.789695.n4.nabble.com/when-to-use-textConnection-tp2327132p2327132.html Sent from the R help mailing list

Hello How do you compare R to SAS in terms of speed and management of large datasets? What about Revolution R? I've seen on their site, they claim that Revolution R is much faster than R and it's multithread... Can you really notice the difference?. What dissadvantage does it have? I think it's based on R 2.10. but R already issued the version 2.12 Regards What alternative

Hello Why this works: ncota <- 1 nslope <- 29 resul <- matrix(rep(0,ncota*nslope*4),ncota*nslope,4) But this doesn't? ncota <- 1 sini <- 0.1; sfin <- 1.5; spaso <- 0.05; nslope <- 1+((sfin-sini)/spaso) resul <- matrix(rep(0,ncota*nslope*4),ncota*nslope,4) I guess the problem is that the division gives a noninteger number. How can I get the second one work? I

Hello I've used read.table to read a file that contains numbers such as 0.00001 when I write them back with write.table those numbers appear as 1e-5 How can I keep the old format? thanks -- View this message in context: http://r.789695.n4.nabble.com/number-format-writing-1e-5-instead-of-0-00001-tp3003831p3003831.html Sent from the R help mailing list archive at Nabble.com.

Hi Usually "aggregate" is used to calculate things such as the sum of all data on the first day, the sum next day, and so on. But how can I calculate the mean of the first hour of all days, the mean of the second hour of all days, and so on. ??? That's Most examples: today at 1am + today at 2am + today at 3am +.... -> sum today tomorrow at 1am + tomorrow at

hello I think I've found a bug I don't know if it's a chron bug or a R one. (05/12/05 23:00:00) +1/24 gives (05/12/05 24:00:00) instead of (05/13/05 00:00:00) it looks like the same but it's not because when you get the date of this datetime it says day 12 instead of 13. Please, forward it to the place where this bugs are supposed to be posted. cheers -- View this message

Hello I have a file with this format 2005-01-03 09:05 0.00 2005-01-03 09:10 0.01 2005-01-03 09:15 0.02 2005-01-03 09:20 0.03 2005-01-03 09:25 0.04 2005-01-03 09:30 0.05 2005-01-03 09:35 0.06 2005-01-03 09:40 0.07 2005-01-03 09:45 0.08 2005-01-03 09:50 0.09 2005-01-03 09:55 0.10 2005-01-03 10:00 0.00 2005-01-03 10:05 0.00 .... some

Hello I have a table of this kind: function x1 x2 x3 2.232 1 1 1.00 2.242 1 1 1.01 2.732 1 1 1.02 2.770 1 2 1.00 1.932 1 2 1.01 2.132 1 2 1.02 3.222 1.2 1 1 ..... ... .. .. The table represents the values of a function(x1, x2, x3) for each combination x1, x2, x3. I'd like to generate a plot where each point has the coordinates x=x1, y=x2,

Hi everybody, I have a small question about the function "na.locf" from the package "zoo". I saw in the help that this function is able to fill NA gaps with the last value before the NA gap (or with the next value). But it is possible to fill my NA gaps according to the last AND the next value at the same time? Actually, I want R to fill my gaps with the method of

I seem to have a (recent) problem with mongrel on NetBSD. I''m running a development release of NetBSD (called NetBSD-current 4.99.34). When I start mongrel, it listens on the IPv6 wildcard address, but not on the IPv4 wildcard: => Booting Mongrel (use ''script/server webrick'' to force WEBrick) => Rails application starting on http://0.0.0.0:3000 => Call with

"Enforce" is overstating it... results will differ if there are no non-NA values for a given ID, and there is a potential further discrepancy if there are multiple non-NA values. But these issues were not identified by the OP, so may not be relevant in their case. -- Sent from my phone. Please excuse my brevity. On January 8, 2018 6:41:33 AM PST, Eric Berger <ericjberger at

Hi R-helpers I'm looking for a vectorised function which does missing value replacement as in last observation carried forward in the zoo package but instead of a locf, I would like the locf function to add +1 to each time a missing value occurred. See below for an example. > require(zoo) > x <- 5:15 > x[4:7] <- NA > coredata(na.locf(zoo(x))) [1] 5 6 7 7 7 7 7 12 13

Yes, you are right if the IDs are always sequentially-adjacent and the first non-NA value appears in the first record for each ID. -- Sent from my phone. Please excuse my brevity. On January 8, 2018 2:29:40 AM PST, PIKAL Petr <petr.pikal at precheza.cz> wrote: >Hi > >With the example, na.locf seems to be the easiest way. >> library(zoo) > >> na.locf(df1) > ID

I?m confused by the difference in the fit of a gam model (in package mgcv) when I specify an interaction in different ways. I would appreciate it if someone could explain the cause of these differences. For example: x <- c(105, 124, 124, 124, 144, 144, 150, 176, 178, 178, 206, 206, 212, 215, 215, 227, 229, 229, 229, 234, 234, 254, 254, 290, 290, 303, 334, 334, 334, 344,